Two moles of helium gas undergo a cyclic process as shown in Fig. Assuming the gas to be ideal, calculate the following quantities in this process

(a) The net change in the heat energy
(b) The net work done
(c) The net change in internal energy

Answer

A, B, C, D

Explanation

Number of moles, `n=2`
Helium is mono-atomic gas, therefore,
`C_(v)=(3)/(2)R , C_(P)=(5)/(2) R`
The gas undergoes cyclic process.
Since, internal energy is proprty of the system, the net change in internal enrgy during the cyclic process is zero.
The net change in the heat energy is equal to the net work done.
`(DeltaQ)_(Net)=(DeltaQ)_(AB)+(DeltaQ_(BC))+(DeltaQ_(CD))+(DeltaQ_(DA))`
`(DeltaQ)_(AB)=nxxC_(P) xx (T_(B)-T_(A))`
`=2xx(5)/(2)xx8.32(400-300)=4160J`
Since Process `BC` is isothermal, therefore `DeltaU=0`
`(DeltaQ)_(BC)=(DeltaW)_(BC)=nRTln(V_(C)/(V_(B)))=nRTln(P_(B)/(P_(C)))`
`=2xx8.32xx400ln ,(2/(1))=4613.6J`
` (DeltaQ_(CD))=nC_(P)(T_(D)-T_(C))`
`=2xx(5)/(2)xx8.32xx(300-400)-4160J`
`(DeltaQ)_(DA)=nRTln , (P_(D)/(P_(A)))`
`=2xx8.32xx300ln, ((1)/(2))=-3460.2J`
`(DeltaW)_(Net) =4160+4613.6-4160-3460.2`
`=1153.4J`